Re: Fireworks math.
- From: AZFIREBALL <AZFIREBALL@xxxxxxxxx>
- Date: Wed, 27 Aug 2008 21:45:43 -0700 (PDT)
On Aug 27, 9:12 pm, "Androcles" <Headmas...@xxxxxxxxxxxxxxxx> wrote:
"AZFIREBALL" <jhohi...@xxxxxxx> wrote in message
news:06f2f0ac-2b55-4920-b81f-253cdeac4613@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Some people, on a pyro site I read, say a ball shot into the air
(fireworks) takes 1/3 of its total flight time going up and 2/3 of
the total flight time coming down.
Can anyone show me the math/theory that supports or disproves this
statement?
Thanks.
Which would you prefer:
a) the theory that supports the conjecture,
b) the theory that denies the conjecture,
c) the evidence of experiment?
It only takes a few minutes for the shuttle to reach orbit where
it may spend a couple of weeks, then maybe an hour or two
to return, but that's not a ball shot into the air.
Where fireworks are concerned the "ball" is usually broken up
to provide a display and the particles will take longer to return,
some of which may be on small parachutes which you will not
see in darkness but will keep a slow burning flare aloft for
quite a long time.
For a heavy cannon ball, the flight time up equals the flight time
down the equation for which is h = 1/2 gt^2.
So are we talking bottle rocket or roman candle, and is there
a starburst at the apex of flight?
I was not clear in my post. The 'ball' is a test object of a given
size and weight that does not explode. It is simply fired out of a gun
(tube) using a known amount of powder and then falls back to earth,
The total flight time is recorded to get some measure of lift powder
effectiveness. This method is used to evaluate the relative lifting
power of different batches of black powder prior to shooting a live
shell.
.
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